Tích \(\left(-3\right)^2.\left(-4\right)\) bằng :
(A) -36 (B) 36 (C) -24 (D) 24
Tìm x:
a) \(\dfrac{x}{4}=\dfrac{4}{x}\)
b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)
c) \(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)
d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
a) \(\dfrac{x}{4}=\dfrac{4}{x}\)=>x2=4.4=16 =>x2=42
=>x=2 hay x=-2.
b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)=>\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)=>x+7=-\(\dfrac{2}{3}.15\)=-10 =>x=-17
c)\(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)=>(x+1)2=2.8=16=42
=>x+1=4 hay x+1=-4
=>x=3 hay x=-5.
d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)=>\(\dfrac{2x-1}{9}=\dfrac{9}{2x-1}\)=>(2x-1)2=92
=>2x-1=9 hay 2x-1=-9
=>x=5 hay x=-4.
Tính các thương sau:
a) \(12:6\)
b) \(24:\left( { - 8} \right)\)
c) \(\left( { - 36} \right):9\)
d) \(\left( { - 14} \right):\left( { - 7} \right)\)
Cách 1:
a) \(12:6 = 2\)
b) \(24:\left( { - 8} \right)=-(24:8)=-3\)
c) \(\left( { - 36} \right):9=-(36:9)=-4\)
d) \(\left( { - 14} \right):\left( { - 7} \right)=14:7=2\)
Cách 2:
a) Ta có \(12 = 6.2\) nên \(12:6 = 2\).
b) Ta có \(24 = \left( { - 8} \right).\left( { - 3} \right)\)\( \Rightarrow 24:\left( { - 8} \right) = \left( { - 3} \right)\).
c) Ta có \(\left( { - 36} \right) = 9.\left( { - 4} \right)\) nên \(\left( { - 36} \right):9 = \left( { - 4} \right)\).
d) Ta có \(\left( { - 14} \right) = \left( { - 7} \right).2\) nên \(\left( { - 14} \right):\left( { - 7} \right) = 2\)
Tính giá trị biểu thức:
a,\(A=\dfrac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)
b,\(^{B=\dfrac{\left[\left(6,2:0,31-\dfrac{5}{6}.0,9\right).0,2+0,15\right]:12}{\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}}}\)
c,\(C=\dfrac{\left(\dfrac{3}{4}\right)^3+\left(\dfrac{5}{4}\right)^3-5\left(\dfrac{4}{3}-\dfrac{5}{4}\right)}{\left(\dfrac{-5}{8}\right)^2+\left(\dfrac{2}{3}\right)^2-\dfrac{5}{6}}\)
d,\(D=\left[\dfrac{\dfrac{17}{24}.9\dfrac{1}{2}-3\dfrac{1}{4}.\dfrac{17}{24}}{3\dfrac{1}{2}.2\dfrac{13}{36}+2\dfrac{13}{36}.2\dfrac{3}{4}}-\dfrac{1}{5}\right]^{-2}\)
Giúp em với chị @Nhã Doanh ới!!!!!!!!
Mai em phải đi thi rồi :((
a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)
\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)
\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)
\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)
\(a\frac{3}{5}-\left(-\frac{1}{2}\right)+\frac{2}{5} b\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}c\left(\frac{3^4}{5}\right).\left(\frac{5^3}{3}\right)d\frac{11}{23}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
Tìm \(\dfrac{\left(\left(4\right)^{-2}:\left(\dfrac{1}{3}\right)^2\right)\dfrac{^1}{2}}{\left(-\dfrac{1}{6}\right)^2}\)
A. 36 B.27 C. 48 D.- 36 E.-27
`(4^{-2}:(1/3)^2xx1/2)/((-1/6)^2)`
`=((1/16:1/9)xx1/2)/(1/36)`
`=36xx1/2xx(1/16xx9)`
`=18xx9/16`
`=81/8`
Thực hiện phép tính:
a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
b)\(-12:\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
c)\(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
d)\(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
e)\(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(0,8-\frac{3}{4}\right)^2\)
a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)
= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)
=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)
=\(\frac{1}{2}-1-1\)
=\(\frac{-3}{2}\)
b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)
= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)
= \(-12:\left\{\frac{-1}{12}\right\}^2\)
= \(-12:\frac{1}{144}\)
= \(-12.144\)
= -1728
c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)
= \(\frac{-7}{6}\)
d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)
= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)
= \(10.\frac{7}{5}\)
= 14
e) (1+23−14).(0,8−34)2
= (1+23−14).(\(\frac{4}{5}\)−34)2
= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
= \(\frac{17}{20}.\frac{1}{400}\)
= \(\frac{17}{8000}\)
phân tích đa thức thành nhân tử theo phương pháp đặt ẩn phụ
\(a.25y^2+10y^8+1\)
\(b.x^4+6x^3+7x^2-6x+1\)
\(c,\frac{36}{x^6}-\frac{24}{x^3}+4\)
\(d,\left(x^2-1\right)-18\left(x+1\right)\left(x-1\right)\)
a)Bạn xem lại đề được không
b)Đặt x^2 ra ngoài
c)Đặt x^3=t rồi quy đồng
d)Bt = -17(x^2-1), còn ẩn phụ gì nữa?
tại thấy thầy ghi đề đặt ẩn phụ nên như vậy,tui cũng nghĩ ra như vậy rùi mà
Bài 2: tính
a, \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
b,\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
c,\(\dfrac{7}{23}.[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}]\)
d,\(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
e,\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-1728\)
c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)
\(=\dfrac{7}{23}.\dfrac{-23}{6}\)
\(=\dfrac{-7}{6}\)
d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)
\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)
\(=10.\dfrac{7}{5}\)
\(=14\)
e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
1.Tính :
a.\(12+\left\{45-\left[36:\left(12-9\right)^2\right]\right\}\)
b.\(24:\left[48-\left(42:7\right)^2\right]\)
2.Viết tập hợp các số tự nhiên không quá 10 bằng hai cách.
Bài 1:
a.\(12+\left\{45-\left[36:\left(12-9\right)^2\right]\right\}\)
\(=12+\left\{45-\left[36:3^2\right]\right\}\)
\(=12+\left\{45-\left[36:9\right]\right\}\)
\(=12+\left\{45-4\right\}\)
\(=12+41\)
\(=53\)
b.\(24:\left[48-\left(42:7\right)^2\right]\)
\(=24:\left[48-6^2\right]\)
\(=24:\left[48-36\right]\)
\(=24:12=2\)
Bài 2 :
C1 : { 0;1;2;3;4;5;6;7;8;9;10}
C2 : \(\left\{x\in N|x\le10\right\}\)
pha ngoac ra la dc
con bai 2 C1 {1:2:3:4:5:6:7:8:9:0}
C2 {nt huoc N /n<10}
a) 12 + {45 - [36 : (12 - 9)2 ]}
= 12 + [45 - (36 : 32)]
= 12 + [45 - (36 : 9)]
= 12 + (45 - 4)
= 12 + 41
= 53